Is the Remainder Term in the Maclaurin Series for Sine Negative?

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Homework Help Overview

The discussion revolves around the Maclaurin series for the sine function, specifically focusing on the remainder term and whether it is negative. Participants are examining the expression for the sine function and its approximation using the series expansion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to prove that the remainder term in the Maclaurin series for sine is negative. There are questions about the use of the Maclaurin series for sine and cosine, as well as inquiries into the nature of the terms following the initial terms of the series.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the remainder term and its implications. Some guidance has been offered regarding the comparison of successive terms in the series, but there is no explicit consensus on the approach to take.

Contextual Notes

There is a focus on proving a specific property of the remainder term, and participants are questioning the assumptions related to the behavior of the series terms. The original poster has indicated a need to show that the remainder is negative, which may imply constraints on the values of x being considered.

transgalactic
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prove that..
[tex] x-\frac{x^3}{6}+\frac{x^5}{120}>\sin x\\[/tex]
[tex] R_5=\frac{f^{5}(c)x^5}{5!}[/tex]
i need to prove that the remainder is negative .
[tex] \sin x=x-\frac{x^3}{6}+\frac{x^5}{120}+R_5[/tex]
[tex] R_5=\frac{cos(c)x^5}{5!}[/tex]
 
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I'm going to guess that you will use the Maclaurin series for sin(x) and cos(x).
 
no?/////

i need to show that the remainder is negative??
 
[tex]x-\frac{x^3}{6}+\frac{x^5}{120}[/tex]
These are the first three terms of the Maclaurin series for sin(x). Do you know what the terms immediately following them look like? Do you know how successive terms in this Maclaurin series compare in absolute value?
 

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